Problem: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $k = \dfrac{12r^2 - 8r}{4r} \times \dfrac{9}{6(3r - 2)} $
Explanation: When multiplying fractions, we multiply the numerators and the denominators. $k = \dfrac{ (12r^2 - 8r) \times 9 } { 4r \times 6(3r - 2) } $ $ k = \dfrac {9 \times 4r(3r - 2)} {4r \times 6(3r - 2)} $ $ k = \dfrac{36r(3r - 2)}{24r(3r - 2)} $ We can cancel the $3r - 2$ so long as $3r - 2 \neq 0$ Therefore $r \neq \dfrac{2}{3}$ $k = \dfrac{36r \cancel{(3r - 2})}{24r \cancel{(3r - 2)}} = \dfrac{36r}{24r} = \dfrac{3}{2} $